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A stream of positively charged particles having $$\frac{q}{m} = 2 \times 10^{11}$$ C kg$$^{-1}$$ and velocity $$\vec{v_0} = 3 \times 10^7 \hat{i}$$ m s$$^{-1}$$ is deflected by an electric field $$1.8\hat{j}$$ kV m$$^{-1}$$. The electric field exists in a region of 10 cm along $$x$$ direction. Due to the electric field, the deflection of the charge particles in the $$y$$ direction is _____ mm.
Correct Answer: 2
Time in field: $$t = 0.1/(3\times10^7) = \frac{1}{3}\times10^{-8}$$ s.
Acceleration: $$a = (q/m)E = 2\times10^{11}\times1800 = 3.6\times10^{14}$$ m/s$$^2$$.
Deflection: $$y = \frac{1}{2}at^2 = 2\times10^{-3}$$ m $$= 2$$ mm.
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